Unit 3: Gravity and electromagnetism

QLDPhysicsSyllabus dot point

Topic 1: Gravity and motion

Apply the relationships for uniform circular motion, including centripetal acceleration a = v^2/r, centripetal force F = m v^2 / r, period T = 2 pi r / v, and the geometry of banked curves and conical pendulums

A focused answer to the QCE Physics Unit 3 dot point on uniform circular motion. Defines centripetal acceleration, identifies the real forces that supply centripetal force in common contexts (string tension, friction, normal-force component, gravity), and works the banked curve and conical pendulum geometries that QCAA expects in IA1 and IA2.

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What this dot point is asking

QCAA wants you to apply the kinematic and dynamic relationships for an object moving in a circle at constant speed, identify the real force (or component of a real force) supplying the centripetal force in a given context, and solve geometry problems for banked curves and conical pendulums. The dot point appears in IA1 short response, often as a banked-curve stimulus, and is a popular IA2 design (conical pendulum or rotating-mass setup).

The answer

Uniform circular motion in one paragraph

An object moving at constant speed vv around a circle of radius rr has changing velocity (because direction is changing). The acceleration is directed toward the centre of the circle (centripetal):

ac=v2ra_c = \frac{v^2}{r}

By Newton's second law, this requires a net inward force:

Fc=mac=mv2rF_c = m a_c = \frac{m v^2}{r}

Centripetal force is not a separate kind of force. It is the role played by whichever real force (or component of one) points toward the centre. In different contexts:

Context Source of IMATH_7
Mass on a string in a horizontal circle Tension (horizontal component for a conical pendulum)
Car on a flat bend Static friction between tyres and road
Car on a banked curve (frictionless) Horizontal component of the normal force
Satellite in orbit Gravitational attraction toward the central body
Charged particle in a magnetic field Magnetic Lorentz force IMATH_8

Period and frequency

For one complete revolution the object travels a distance 2πr2 \pi r at speed vv, so the period is:

T=2πrvT = \frac{2 \pi r}{v}

and the frequency is f=1/Tf = 1/T. Angular speed is ω=2πf=v/r\omega = 2 \pi f = v/r, so ac=ω2ra_c = \omega^2 r is an equivalent form.

Banked curves

On a curve banked at angle θ\theta to the horizontal, the normal force is perpendicular to the road surface. Resolving:

  • Vertical: Ncosθ=mgN \cos\theta = m g (no vertical acceleration).
  • Horizontal: Nsinθ=mv2/rN \sin\theta = m v^2 / r (centripetal).

Dividing eliminates NN and mm:

tanθ=v2rg\tan\theta = \frac{v^2}{r g}

This is the design speed of the banked curve. At that speed, no friction is required to keep the vehicle on the curve. Above the design speed, friction must act down the slope (toward the inside of the curve); below it, friction must act up the slope.

Try it: Banking angle calculator. Enter speed and radius to get the design angle and the friction required at any other speed.

Conical pendulum

A mass swung in a horizontal circle on a string at angle θ\theta from the vertical has tension components:

  • Vertical: Tcosθ=mgT \cos\theta = m g.
  • Horizontal: Tsinθ=mv2/rT \sin\theta = m v^2 / r.

The relationship between angle and speed is identical to the banked curve: tanθ=v2/(rg)\tan\theta = v^2 / (r g). The radius of the horizontal circle is r=Lsinθr = L \sin\theta where LL is the string length.

The period of a conical pendulum:

T=2πLcosθgT = 2 \pi \sqrt{\frac{L \cos\theta}{g}}

A classic IA2 measures period as a function of cone angle (or string length) and linearises against cosθ\cos\theta (or LL) to extract gg.

Worked example: car on a flat bend

A 900 kg car rounds a flat bend of radius 30 m at 12 m/s. The required centripetal force is Fc=900×122/30=4320F_c = 900 \times 12^2 / 30 = 4320 N, supplied entirely by static friction.

Try it: Centripetal force calculator. Enter mass, speed and radius for FcF_c and aca_c.

How this appears in IA1 and IA2

IA1 data test. Expect a banked-curve diagram with the bank angle and a speed, asked to determine whether the vehicle stays on the curve, to find the design speed, or to extract the friction coefficient required at a non-design speed. Alternatively, a conical pendulum stimulus with multiple period measurements at different angles.

IA2 student experiment. Common designs: a conical pendulum with mass, period and angle measured to test T=2πLcosθ/gT = 2 \pi \sqrt{L \cos\theta / g}; a rotating-mass setup on a horizontal turntable with friction varied; a ball-in-a-cone or marble-in-a-bowl rolling at different heights. Strong reports linearise the relationship before fitting (e.g. T2T^2 vs cosθ\cos\theta slopes give 4π2L/g4\pi^2 L / g) and propagate uncertainty cleanly.

Common traps

Treating centripetal force as a new force. It is the inward-pointing role played by an existing real force. Always state which real force (tension, friction, gravity, normal-component) supplies it.

Pointing FcF_c tangentially or outward. Centripetal force is always directed toward the centre of the circle. The "centrifugal" force students sometimes invoke is a pseudo-force that appears only in the rotating frame; not used in QCAA solutions.

Mixing up vv, ω\omega and ff. v=ωrv = \omega r and ω=2πf\omega = 2 \pi f. Substitute consistently.

Banking with mass dependence. The design angle tanθ=v2/(rg)\tan\theta = v^2 / (r g) does not depend on mm. If mm appears in your final answer for the bank angle, you have made an error.

Treating uniform circular motion as zero acceleration. Constant speed does not mean constant velocity. The acceleration is non-zero because the direction is changing.

In one sentence

In uniform circular motion an object travels at constant speed vv around radius rr with centripetal acceleration v2/rv^2/r directed toward the centre, supplied by whichever real force points inward (tension, friction, normal-component, gravity, Lorentz), and banked-curve and conical-pendulum geometries both reduce to tanθ=v2/(rg)\tan\theta = v^2 / (r g).

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

2023 QCAA-style6 marksA 1200 kg car drives around a flat (unbanked) circular bend of radius 45 m at 14 m/s. (a) Calculate the centripetal acceleration and the centripetal force required. (b) State which real force supplies this. (c) Calculate the minimum coefficient of static friction required between the tyres and the road. (d) If the road is now banked at an angle so that no friction is required at 14 m/s, calculate this angle. Use g = 9.8 m/s^2.
Show worked answer →

A 6-mark answer needs the centripetal kinematics, the force identification, the friction calculation, and the bank-angle calculation.

(a) Centripetal acceleration and force.

ac=v2/r=142/45=4.36 m/s2a_c = v^2 / r = 14^2 / 45 = 4.36 \text{ m/s}^2.
Fc=mac=1200×4.36=5230 NF_c = m a_c = 1200 \times 4.36 = 5230 \text{ N} (directed horizontally toward the centre).

(b) Source of the force. On a flat road, the only horizontal force on the car is the static friction between the tyres and the road, directed toward the centre of the bend.

(c) Minimum coefficient. Static friction must supply the entire FcF_c:

μsmgmv2/r\mu_s m g \geq m v^2 / r, so μsv2/(rg)=4.36/9.8=0.445\mu_s \geq v^2 / (r g) = 4.36 / 9.8 = 0.445.

(d) Bank angle for no friction. Resolve the normal force into a vertical component balancing gravity (Ncosθ=mgN \cos\theta = m g) and a horizontal component supplying centripetal force (Nsinθ=mv2/rN \sin\theta = m v^2 / r). Dividing:

tanθ=v2/(rg)=142/(45×9.8)=0.445\tan\theta = v^2 / (r g) = 14^2 / (45 \times 9.8) = 0.445, so θ=24.0°\theta = 24.0°.

Markers reward the explicit centre-pointing direction of FcF_c, the identification of friction as the supplying force, the inequality on μs\mu_s, and the derivation of tanθ\tan\theta from the vertical and horizontal force balances.

2022 QCAA-style4 marksA 0.50 kg mass is swung in a horizontal circle of radius 0.80 m on a string that makes an angle of 30 degrees with the vertical (a conical pendulum). Calculate the tension in the string and the speed of the mass. Use g = 9.8 m/s^2.
Show worked answer →

The string tension TT has a vertical component Tcos30°T \cos 30° balancing gravity and a horizontal component Tsin30°T \sin 30° supplying centripetal force toward the vertical axis of the cone.

Tension. From the vertical balance:

Tcos30°=mgT=(0.50×9.8)/0.866=5.66 NT \cos 30° = m g \Rightarrow T = (0.50 \times 9.8) / 0.866 = 5.66 \text{ N}.

Speed. Horizontal component supplies centripetal force:

IMATH_4
IMATH_5
IMATH_6
v=2.13 m/sv = 2.13 \text{ m/s}.

Markers reward identifying the two components of tension and their roles, the vertical balance giving TT, and the horizontal balance giving vv, with consistent units throughout.

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